Question

answer the question guys​

Answer 1

Answer:

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Answer 2

Answer:

$\huge {\red {\mathtt {\boxed {Given:}}}}$

$\large {{7}^{96} - {3}^{58}}$

$\huge {\green {\mathtt {\boxed {Solution:}}}}$

First term = $\large{7}^{96}$

$\large{7}^{1}$ = $7$

$\large{7}^{2}$ = $49$

$\large{7}^{3}$ = $343$

$\large{7}^{4}$ = $2401$

$\large{7}^{5}$ = $16807$

So, it is observed, that the last digits repeat as 7, 9, 3, 1, 7, ...

Therefore, number of possible unit cycles = 4

Which means there can be 4 types of unit digits i.e., 7, 9, 3, and 1 for all powers of 7.

Now, we can compute the last digit of a larger power of number by taking out the modulo 4 of the power.

(Modulo- It is the remainder of a number when divided by divisor)

Therefore,

=$\large{4 \: mod \: 96 = 0}$

Here, the remainder is 0, and 7⁰ = 1

Hence, the unit digit of $\large {{7}^{96} = 1}$

__________________________________

Now, in the same way, we get that the unit cycle of any power of 3 is also 4.

Therefore,

${66 \: mod \: 4 = 2}$

and,

$\large {{3}^{2} = 9}$,

Hence, the unit digit of $\large {{3}^{66} = 9}$

Now,

$\large {{7}^{96} - {3}^{58}}$

= $9 - 1$

= $-8$

= absolute value of -8 which is 8 (Ans)

$\huge {\red {\mathtt {\boxed {Answer:}}}}$

$\large {Option D}$

Thanks for posting such a nice question, happy to solve it. Hope you liked my answer, cheers :)